Summary

You searched for: inst=220680

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1

New Number: 5.100 |  AESZ: 347  |  Superseeker: 15 27140/3  |  Hash: f00de20026c099e75b447c475ab287e4  

Degree: 5

\(\theta^4-3 x\left(213\theta^4+186\theta^3+149\theta^2+56\theta+8\right)+2^{3} 3^{3} x^{2}\left(702\theta^4+1078\theta^3+949\theta^2+392\theta+60\right)-2^{6} 3^{3} x^{3}\left(9277\theta^4+18432\theta^3+16008\theta^2+6000\theta+840\right)+2^{13} 3^{4} 5 x^{4}(2\theta+1)^2(51\theta^2+69\theta+32)-2^{14} 3^{6} 5^{2} x^{5}(2\theta+1)^2(2\theta+3)^2\)

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Coefficients of the holomorphic solution: 1, 24, 1944, 218400, 28488600, ...
--> OEIS
Normalized instanton numbers (n0=1): 15, 1329/4, 27140/3, 220680, 5952570, ... ; Common denominator:...

Discriminant

\(-(192z-1)(1728z^2-207z+1)(-1+120z)^2\)

Local exponents

\(0\)\(\frac{ 23}{ 384}-\frac{ 11}{ 1152}\sqrt{ 33}\)\(\frac{ 1}{ 192}\)\(\frac{ 1}{ 120}\)\(\frac{ 23}{ 384}+\frac{ 11}{ 1152}\sqrt{ 33}\)\(\infty\)
\(0\)\(0\)\(0\)\(0\)\(0\)\(\frac{ 1}{ 2}\)
\(0\)\(1\)\(1\)\(1\)\(1\)\(\frac{ 1}{ 2}\)
\(0\)\(1\)\(1\)\(3\)\(1\)\(\frac{ 3}{ 2}\)
\(0\)\(2\)\(2\)\(4\)\(2\)\(\frac{ 3}{ 2}\)

Note:

This is operator "5.100" from ...

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